To solve the integral $\int \tan^4(5x) \, dx$, we can rewrite $\tan^4(5x)$ using trigonometric identities to simplify the expression.

Step 1: Express $\tan^4(5x)$ in terms of $\tan^2(5x)$

We use the identity: $\tan^2(\theta) = \sec^2(\theta) - 1$ Thus, $\tan^4(5x)$ can be written as: $\tan^4(5x) = (\tan^2(5x))^2 = \left(\sec^2(5x) - 1\right)^2$ Expanding the square: $\tan^4(5x) = \sec^4(5x) - 2\sec^2(5x) + 1$

Step 2: Split the integral

Now, we can split the integral into three simpler integrals: $\int \tan^4(5x) \, dx = \int \left( \sec^4(5x) - 2\sec^2(5x) + 1 \right) dx$ This becomes: $\int \sec^4(5x) \, dx - 2\int \sec^2(5x) \, dx + \int 1 \, dx$

Step 3: Solve each integral

1. Integral of $\sec^4(5x)$:

   Use the reduction formula for powers of secant: $\int \sec^4(5x) \, dx = \frac{1}{3} \tan(5x) \sec^2(5x) + \frac{2}{15} \tan^3(5x) + C_1$

2. Integral of $\sec^2(5x)$:

   This is a standard integral: $\int \sec^2(5x) \, dx = \frac{1}{5} \tan(5x)$

3. Integral of 1: $\int 1 \, dx = x$

Step 4: Combine the results

Putting everything together: $\int \tan^4(5x) \, dx = \frac{1}{3} \tan(5x) \sec^2(5x) + \frac{2}{15} \tan^3(5x) - \frac{2}{5} \tan(5x) + x + C$ where $C$ is the constant of integration.

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Would you like more details on any step?

Here are 5 related questions you might find interesting:

1. How would you solve $\int \sec^n(x) \, dx$ for general $n$?
2. Can you derive the reduction formula for $\sec^n(x)$?
3. What is the integral of $\tan^n(x)$ for general $n$?
4. How do trigonometric identities simplify complex integrals?
5. How would you approach the integral of $\cot^4(x)$?

Tip: Always look for identities to simplify trigonometric integrals before starting to integrate!